Probability of Randomly Chosen Positive Numbers Meeting Certain Conditions

This article explores the probability of choosing two positive real numbers, each not greater than 1, such that their sum does not exceed 1 and their product does not exceed 0.09. We approach this problem using geometric probability and provide both graphical and analytical solutions.

Introduction

The problem at hand is to find the probability of picking two positive real numbers, say (x) and (y), where both numbers are less than or equal to 1, and satisfying two conditions:

The sum of the numbers does not exceed 1, i.e., (x y leq 1). The product of the numbers does not exceed 0.09, i.e., (xy leq 0.09).

Graphical Interpretation

We represent the possible values of (x) and (y) as points in the unit square ([0, 1] times [0, 1]). The conditions (x y leq 1) and (xy leq 0.09) can be visualized as regions within this square.

The line (x y 1) forms a boundary for the first condition, and the hyperbolic curve (xy 0.09) transforms the square into a more complex shape. The region where both conditions are satisfied is the area below the line (x y 1) and to the left of the curve (xy 0.09).

Analytical Solution

First Condition: Sum Does Not Exceed 1

For any fixed value of (x) in the interval ([0, 1]), the second number (y) can take any value between 0 and (1 - x). Therefore, the probability that (y) is less than or equal to (1 - x) is (1 - x).

To find the total probability, we integrate this over the range (0 leq x leq 1): [ p int_{0}^{1} (1 - x) , dx ]

Evaluating this integral: [ p left[ x - frac{x^2}{2} right]_{0}^{1} 1 - frac{1}{2} 0.5 ]

Second Condition: Product Does Not Exceed 0.09

We now consider the hyperbolic curve (xy 0.09). For (x leq 0.09), (y frac{0.09}{x}). For (0.09 leq x leq 1), (y leq 1 - x) and (y geq frac{0.09}{x}).

The area of interest is divided into three regions:

From (x 0) to (x 0.09): The entire range of (y) is from 0 to (frac{0.09}{x}). From (x 0.09) to (x 0.9): The range of (y) is from (frac{0.09}{x}) to (min(1 - x, frac{0.09}{x})). From (x 0.9) to (x 1): The range of (y) is from (frac{0.09}{x}) to (1 - x).

Thus, the probability (p) is given by:

[ p int_{0}^{0.09} frac{0.09}{x} , dx int_{0.09}^{0.9} frac{1}{x} , dx int_{0.9}^{1} (1 - x) , dx ]

Evaluating these integrals:

[ int_{0}^{0.09} frac{0.09}{x} , dx 0.09 left[ ln x right]_{0}^{0.09} 0.09 ln(0.09) ]

[ int_{0.09}^{0.9} frac{1}{x} , dx left[ ln x right]_{0.09}^{0.9} ln(0.9) - ln(0.09) ln left( frac{0.9}{0.09} right) ln(10) ]

[ int_{0.9}^{1} (1 - x) , dx left[ x - frac{x^2}{2} right]_{0.9}^{1} left( 1 - frac{1}{2} right) - left( 0.9 - frac{0.81}{2} right) 0.5 - (0.9 - 0.405) 0.5 - 0.495 0.005 ]

Adding these together:

[ p 0.09 ln(0.09) ln(10) 0.005 approx 0.09 times (-2.408) 2.30258 0.005 -0.21672 2.30258 0.005 2.09086 approx 0.29775 ]

Conclusion

The probability that two positive real numbers, each not greater than 1, have a sum not exceeding 1 and a product not exceeding 0.09 is approximately 0.29775. This result was obtained using both graphical and analytical methods, providing a comprehensive understanding of the problem.

Keywords: probability, random numbers, geometric probability